Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 21, Problem 21.18P

(a)

Interpretation Introduction

Interpretation:

The given skeleton reaction has to be balanced.  The reducing and the oxidizing agents has to be identified.

As4O6(s)+MnO4(aq)AsO43(aq)+Mn2+(aq)[Acidic]

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

Anode: The electrode where the oxidation occurs is called as an anode.  It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode.  It is a positively charged electrode.

Steps in balancing organic redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.
  11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Reducing agent: The substance that reduces another substance and gets oxidized is called as a reducing agent.

Oxidizing agent: The substance that oxidizes another substance and gets reduced is called as an oxidizing agent.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given cell reaction is balanced as follows,

The atoms other than oxygen and hydrogen are balanced first.  Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+.  Finally, the charges are balanced using electrons.

As4O6(s)+MnO4(aq)AsO43(aq)+Mn2+(aq)[Acidic]

Examining the above equation it is clear that reduction of Mn and oxidation of As are taking place.

The two half- reaction:

Reduction half-reaction:

As4O6(s)AsO43(aq)

Oxidation half-reaction:

MnO4(aq)Mn2+(aq)

BalancereductionhalfreactionBalanceoxidationhalfreactionMnO4(aq)Mn2+(aq)MnO4(aq)Mn2+(aq)MnO4(aq)Mn2+(aq)+4H2O(l)MnO4(aq)+8H+(aq)Mn2+(aq)+4H2O(l)MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)As4O6(s)AsO43(aq)As4O6(s)4AsO43(aq)As4O6(s)+10H2O(l)4AsO43(aq)As4O6(s)+10H2O(l)4AsO43(aq)+20H+(aq)As4O6(s)+10H2O(l)4AsO43(aq)+20H+(aq)+8e

Multiply reduction half-reaction by 8 and oxidation half-reaction by 5 to balance the no. of electrons involved in both the reaction.  Then add both the reactions and cancel H2OandH+.

8MnO4(aq)+64H+(aq)+40e8Mn2+(aq)+32H2O(l)5As4O6(s)+50H2O(l)20AsO43(aq)+100H+(aq)+40e_8MnO4(aq)+5As4O6(s)+18H2O(l)8Mn2+(aq)+20AsO43(aq)+36H+(aq)

Hence,

Oxidizing agent: MnO4

Reducing agent: As4O6

(b)

Interpretation Introduction

Interpretation:

The given skeleton reaction has to be balanced.  The reducing and the oxidizing agents has to be identified.

P4(s)HPO32(aq)+PH3(g)[Acidic]

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

Anode: The electrode where the oxidation occurs is called as an anode.  It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode.  It is a positively charged electrode.

Steps in balancing organic redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.
  11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Reducing agent: The substance that reduces another substance and gets oxidized is called as a reducing agent.

Oxidizing agent: The substance that oxidizes another substance and gets reduced is called as an oxidizing agent.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given cell reaction is balanced as follows,

The atoms other than oxygen and hydrogen are balanced first.  Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+.  Finally, the charges are balanced using electrons.

P4(s)HPO32(aq)+PH3(g)[Acidic]

Since both products contain phosphorus, divide the half-reactions so each includes P4 as the reactant.

The two half- reaction:

Reduction half-reaction:

P4(s)PH3(aq)

Oxidation half-reaction:

P4(s)4HPO32(aq)

BalancereductionhalfreactionBalanceoxidationhalfreactionP4(s)PH3(g)P4(s)4PH3(g)P4(s)+12H+(aq)4PH3(g)P4(s)+12H+(aq)+12e4PH3(g)P4(s)HPO32(aq)P4(s)4HPO32(aq)P4(s)+12H2O(l)4HPO32(aq)+20H+(aq)P4(s)+12H2O(l)4HPO32(aq)+20H+(aq)+12e

Add both the reactions and cancel H+.

P4(s)+12H+(aq)+12e4PH3(g)P4(s)+12H2O(l)4HPO32(aq)+20H+(aq)+12e_2P4(s)+12H2O(l)4PH3(g)+4HPO32(aq)+8H+(aq)orP4(s)+6H2O(l)2PH3(g)+2HPO32(aq)+4H+(aq)

Hence,

Oxidizing agent: P4

Reducing agent: P4

(c)

Interpretation Introduction

Interpretation:

The given skeleton reaction has to be balanced. The reducing and the oxidizing agents has to be identified.

MnO4(aq)+CN(aq)MnO2(s)+CNO(aq)[Basic]

Concept Introduction:

Oxidation half-reaction: An oxidation half-reaction is a part of redox reaction that shows only for the oxidized species with electrons and its oxidation state increases with the loss of electrons.

Reduction half-reaction: A reduction half-reaction is a part of redox reaction (counterpart of oxidation half-reaction) that shows only for the reduced species with electrons and its oxidation state decreases with the gain of electrons.

Anode: The electrode where the oxidation occurs is called as an anode.  It is a negatively charged electrode.

Cathode: The electrode where reduction occurs is called as a cathode.  It is a positively charged electrode.

Steps in balancing organic redox reactions:

  1. 1) Divide the overall reaction into an oxidation half-reaction and a reduction half-reaction
  2. 2) Balance atoms other than O and H.
  3. 3) Balance O by adding H2O as needed
  4. 4) Balance H by adding H+ ion at the required side.
  5. 5) Balance charges by adding, as needed number of electrons,
  6. 6) Multiply the oxidation half-reaction with the coefficient of electrons in the reduction part.
  7. 7) Multiply the reduction half-reaction with the coefficient of electrons in the oxidation part.
  8. 8) Combine the two half-reactions, cancel out the species that appears on both side, so that number of elements that appear on both sides become equal.
  9. 9) For the reaction in acidic medium, the presence of H+ ion allowed.
  10. 10) For the reaction in basic medium, the H+ ion should be neutralized by the addition of required amount of OH ions on both the sides.
  11. 11) Cancel out the species that appears on both sides and ensure that the number of atoms on the reactant side is equal to the number of atoms on the product side.

Reducing agent: The substance that reduces another substance and gets oxidized is called as a reducing agent.

Oxidizing agent: The substance that oxidizes another substance and gets reduced is called as an oxidizing agent.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given cell reaction is balanced as follows,

The atoms other than oxygen and hydrogen are balanced first.  Then the oxygen atoms are balanced using water and the hydrogen atoms are balanced using H+.  Finally, the charges are balanced using electrons.

MnO4(aq)+CN(aq)MnO2(s)+CNO(aq)[Basic]

Examining the above equation it is clear that reduction of Mn and oxidation of CN- are taking place.

The two half- reaction:

Reduction half-reaction:

MnO4(aq)MnO2(s)

Oxidation half-reaction:

CN(aq)CNO(aq)

BalancereductionhalfreactionBalanceoxidationhalfreactionMnO4(aq)MnO2(s)MnO4(aq)MnO2(s)+2H2O(l)MnO4(aq)+4H+(aq)MnO2(s)+2H2O(l)MnO4(aq)+4H+(aq)+3eMnO2(s)+2H2O(l)CN(aq)CNO(aq)CN(aq)+H2O(l)CNO(aq)CN(aq)+H2O(l)CNO(aq)+2H+(aq)CN(aq)+H2O(l)CNO(aq)+2H+(aq)+2e

Multiply reduction half-reaction by 2 and oxidation half-reaction by 3 to balance the no. of electrons involved in both the reaction.  Then add both the reactions and cancel H2OandH+.

2MnO4(aq)+8H+(aq)+6e2MnO2(s)+4H2O(l)3CN(aq)+3H2O3CNO(aq)+6H+(aq)+6e_2MnO4(aq)+3CN(aq)+2H+(aq)2MnO2(s)+3CNO(aq)+H2O(l)

Add 2 OH to both sides to neutralize H+ and form H2O:

2MnO4(aq)+3CN(aq)+2H+(aq)+2OH(aq)2MnO2(s)+3CNO(aq)+H2O(l)+2OH(aq)

Or

2MnO4(aq)+3CN(aq)+2H2O(l)2MnO2(s)+3CNO(aq)+H2O(l)+2OH(aq)

The balanced chemical reaction in basic medium is

2MnO4(aq)+3CN(aq)+H2O(l)2MnO2(s)+3CNO(aq)+2OH(aq)

Hence,

Oxidizing agent: MnO4

Reducing agent: CN

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Chapter 21 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 21.4 - Prob. 21.6AFPCh. 21.4 - Prob. 21.6BFPCh. 21.4 - Prob. 21.7AFPCh. 21.4 - Prob. 21.7BFPCh. 21.4 - Prob. 21.8AFPCh. 21.4 - Prob. 21.8BFPCh. 21.7 - The most ionic and least ionic of the common...Ch. 21.7 - Prob. 21.9BFPCh. 21.7 - Prob. 21.10AFPCh. 21.7 - Prob. 21.10BFPCh. 21.7 - Prob. 21.11AFPCh. 21.7 - Prob. 21.11BFPCh. 21.7 - In the final steps of the ETC, iron and copper...Ch. 21.7 - Prob. B21.2PCh. 21 - Prob. 21.1PCh. 21 - Prob. 21.2PCh. 21 - Prob. 21.3PCh. 21 - Water is used to balance O atoms in the...Ch. 21 - Prob. 21.5PCh. 21 - Prob. 21.6PCh. 21 - Prob. 21.7PCh. 21 - Prob. 21.8PCh. 21 - Prob. 21.9PCh. 21 - Prob. 21.10PCh. 21 - Prob. 21.11PCh. 21 - Prob. 21.12PCh. 21 - Prob. 21.13PCh. 21 - Prob. 21.14PCh. 21 - Prob. 21.15PCh. 21 - Prob. 21.16PCh. 21 - Prob. 21.17PCh. 21 - Prob. 21.18PCh. 21 - Prob. 21.19PCh. 21 - Prob. 21.20PCh. 21 - Aqua regia, a mixture of concentrated HNO3 and...Ch. 21 - Consider the following general voltaic...Ch. 21 - Why does a voltaic cell not operate unless the two...Ch. 21 - Prob. 21.24PCh. 21 - Prob. 21.25PCh. 21 - Prob. 21.26PCh. 21 - Consider the following voltaic cell: In which...Ch. 21 - Consider the following voltaic cell: In which...Ch. 21 - Prob. 21.29PCh. 21 - Prob. 21.30PCh. 21 - A voltaic cell is constructed with an Fe/Fe2+...Ch. 21 - Prob. 21.32PCh. 21 - Prob. 21.33PCh. 21 - Prob. 21.34PCh. 21 - Prob. 21.35PCh. 21 - What does a negative indicate about a redox...Ch. 21 - Prob. 21.37PCh. 21 - In basic solution, Se2− and ions react...Ch. 21 - Prob. 21.39PCh. 21 - Prob. 21.40PCh. 21 - Use the emf series (Appendix D) to arrange each...Ch. 21 - Prob. 21.42PCh. 21 - Prob. 21.43PCh. 21 - Prob. 21.44PCh. 21 - Prob. 21.45PCh. 21 - Prob. 21.46PCh. 21 - Prob. 21.47PCh. 21 - Prob. 21.48PCh. 21 - Prob. 21.49PCh. 21 - Prob. 21.50PCh. 21 - Prob. 21.51PCh. 21 - Prob. 21.52PCh. 21 - Prob. 21.53PCh. 21 - Prob. 21.54PCh. 21 - Prob. 21.55PCh. 21 - Prob. 21.56PCh. 21 - Prob. 21.57PCh. 21 - Prob. 21.58PCh. 21 - Prob. 21.59PCh. 21 - Prob. 21.60PCh. 21 - Prob. 21.61PCh. 21 - Prob. 21.62PCh. 21 - Prob. 21.63PCh. 21 - Prob. 21.64PCh. 21 - Prob. 21.65PCh. 21 - Prob. 21.66PCh. 21 - Prob. 21.67PCh. 21 - Prob. 21.68PCh. 21 - Prob. 21.69PCh. 21 - Prob. 21.70PCh. 21 - Prob. 21.71PCh. 21 - Prob. 21.72PCh. 21 - Prob. 21.73PCh. 21 - Prob. 21.74PCh. 21 - Prob. 21.75PCh. 21 - Prob. 21.76PCh. 21 - Prob. 21.77PCh. 21 - Prob. 21.78PCh. 21 - Prob. 21.79PCh. 21 - Prob. 21.80PCh. 21 - Prob. 21.81PCh. 21 - Consider the following general electrolytic...Ch. 21 - Prob. 21.83PCh. 21 - Prob. 21.84PCh. 21 - Prob. 21.85PCh. 21 - Prob. 21.86PCh. 21 - In the electrolysis of molten NaBr: What product...Ch. 21 - Prob. 21.88PCh. 21 - Prob. 21.89PCh. 21 - Prob. 21.90PCh. 21 - Prob. 21.91PCh. 21 - Prob. 21.92PCh. 21 - Prob. 21.93PCh. 21 - Prob. 21.94PCh. 21 - Prob. 21.95PCh. 21 - Prob. 21.96PCh. 21 - Prob. 21.97PCh. 21 - Write a balanced half-reaction for the product...Ch. 21 - Prob. 21.99PCh. 21 - Prob. 21.100PCh. 21 - Prob. 21.101PCh. 21 - Prob. 21.102PCh. 21 - Prob. 21.103PCh. 21 - Prob. 21.104PCh. 21 - Prob. 21.105PCh. 21 - Prob. 21.106PCh. 21 - Prob. 21.107PCh. 21 - Prob. 21.108PCh. 21 - Prob. 21.109PCh. 21 - Prob. 21.110PCh. 21 - Prob. 21.111PCh. 21 - Prob. 21.112PCh. 21 - Prob. 21.113PCh. 21 - Prob. 21.114PCh. 21 - Prob. 21.115PCh. 21 - Prob. 21.116PCh. 21 - Prob. 21.117PCh. 21 - Prob. 21.118PCh. 21 - Prob. 21.119PCh. 21 - Prob. 21.120PCh. 21 - To examine the effect of ion removal on cell...Ch. 21 - Prob. 21.122PCh. 21 - Prob. 21.123PCh. 21 - Prob. 21.124PCh. 21 - Prob. 21.125PCh. 21 - Prob. 21.126PCh. 21 - Commercial electrolytic cells for producing...Ch. 21 - Prob. 21.129PCh. 21 - Prob. 21.130PCh. 21 - The following reactions are used in...Ch. 21 - Prob. 21.132PCh. 21 - Prob. 21.133PCh. 21 - Prob. 21.134PCh. 21 - Prob. 21.135PCh. 21 - If the Ecell of the following cell is 0.915 V,...Ch. 21 - Prob. 21.137PCh. 21 - Prob. 21.138PCh. 21 - Prob. 21.139PCh. 21 - Prob. 21.140PCh. 21 - Prob. 21.141PCh. 21 - Prob. 21.142PCh. 21 - Prob. 21.143PCh. 21 - Prob. 21.144PCh. 21 - Prob. 21.145PCh. 21 - Prob. 21.146PCh. 21 - Prob. 21.147PCh. 21 - Both Ti and V are reactive enough to displace H2...Ch. 21 - For the reaction ∆G° = 87.8 kJ/mol Identity the...Ch. 21 - Two voltaic cells are to be joined so that one...Ch. 21 - Prob. 21.152PCh. 21 - Prob. 21.153P
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